# Solve below somewhat symmetric equations: x, y, z subject to x^2+y^2 -

Solve below somewhat symmetric equations:
x, y, z subject to
${x}^{2}+{y}^{2}-xy=3$
${\left(x-z\right)}^{2}+{\left(y-z\right)}^{2}-\left(y-z\right)\left(x-z\right)=4$
${\left(x-z\right)}^{2}+{y}^{2}-y\left(x-z\right)=1$
$x,y,z\in {R}^{+}$
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Volsa280

Step 1
We can take a quadrilateral ABCD, for which:
and $CD=\sqrt{3}$
${x}^{2}+{y}^{2}-xy=E{D}^{2}+E{C}^{2}-2ED\cdot EC{\mathrm{cos}60}^{\circ }=C{D}^{2}=3,$
${\left(x-z\right)}^{2}+{\left(y-z\right)}^{2}-\left(y-z\right)\left(x-z\right)={\left(z-x\right)}^{2}+{\left(z-y\right)}^{2}-\left(z-x\right)\left(z-y\right)=$
$=E{B}^{2}+E{A}^{2}-2EB\cdot EA{\mathrm{cos}60}^{\circ }=A{B}^{2}=4$
and${\left(x-z\right)}^{2}+{y}^{2}-y\left(x-z\right)={\left(z-x\right)}^{2}+{y}^{2}+y\left(z-x\right)=$
$=E{B}^{2}+E{C}^{2}-2EB\cdot EC{\mathrm{cos}120}^{\circ }=1.$
Now, let BFCD be a parallelogram.
Thus,
$AC=z=BD=FC$
and
$\measuredangle ACF=\measuredangle DEC=60°,$
which gives that $\mathrm{△}AFC$ is equilateral triangle,  adn $BC=1$
Now, let R be a rotation around C by $-{60}^{\circ }$ and $R\left(\left\{B\right\}\right)=\left\{G\right\}$ thus, since $\mathrm{△}BGC$ is an equilateral triangle, we obtain:  and
$F{G}^{2}={2}^{2}={\left(\sqrt{3}\right)}^{2}+{1}^{2}=B{F}^{2}+B{C}^{2},$
which says
$\measuredangle FBG=90°,$
$\measuredangle FBC=\measuredangle FBC+\measuredangle GBC=90°+60°=150°,$
which gives
${z}^{2}=A{C}^{2}=F{C}^{2}=F{B}^{2}+B{C}^{2}-2FB\cdot BC{\mathrm{cos}150}^{\circ }=3+1+3=7,$
whisch gives
$z=\sqrt{7}$
and from here we get the answer:
$\left(x,y,z\right)=\left(\frac{5}{\sqrt{7}},\frac{1}{\sqrt{7}},\sqrt{7}\right).$

###### Not exactly what you’re looking for?
Felicity Carter
Step 1
We multiply the thrid equation by four, subtract it from two, and then substitute with equation one solved for x.
$\left[{\left(x-z\right)}^{2}+{\left(y-z\right)}^{2}-\left(y-z\right)\left(x-z\right)\right]$
$-4\left[{\left(x-z\right)}^{2}+{y}^{2}-y\left(x-z\right)\right]=0$
$x=\frac{\sqrt{3}\sqrt{4-{y}^{2}}+y}{2}$
$⇒y=\frac{14-3\sqrt{3}}{13}$
$z=±\sqrt{3}$
It should be easy to figure out x from here.